Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{6z}{z - 1} \times \dfrac{5z - 5}{3} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ 6z \times (5z - 5) } { (z - 1) \times 3 } $ $ x = \dfrac {6z \times 5(z - 1)} {3 (z - 1)} $ $ x = \dfrac{30z(z - 1)}{3(z - 1)} $ We can cancel the $z - 1$ so long as $z - 1 \neq 0$ Therefore $z \neq 1$ $x = \dfrac{30z \cancel{(z - 1})}{3 \cancel{(z - 1)}} = \dfrac{30z}{3} = 10z $